How do I prove that if matrix $A_{n \times n}$ is a skew-symmetric matrix where n is an even number, then A is invertible?

Question

I can prove that if matrix $A_{n \times n}$ is a skew-symmetric matrix where $n$ is an element of the set of odd numbers, then $A$ is not invertible, but I can't prove how $A_{n \times n}$ (skewed symmetric matrix) is invertible if $n$ is an even number.

Answer 1

You cannot prove it because it is not true.

The zero matrix is skew symmetric, and it is not invertible when $n$ is even.

Answer 2

$A=(a_{ij})_{n\times n}$ is invertable, if $\det A \neq 0$

$A=(a_{ij})_{n\times n}$ is a skew-symmetric matrix, so $A^T=-A$ i.e., if $a_{ij}=-a_{ji}$

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For a $n \times n$ matrix $A$,

$1.$ $\det (A) = \det (A^T)$ and

$2.$ $\det (cA)=c^n \det A$

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So here $\det (M)=\det (M^T)=\det (-M)=(-1)^n \det (M)$, as $M=(m_{ij})_{n\times n}$ is a Skew-symmetric matrix.

If the dimension of a skew-symmetric matrix is even i.e., if $n$ is even, then $$\det (M)=\det (M^T)=\det (-M)=(-1)^n \det (M)=\det (M)$$ so we do not have any conclusion about $\det(M)$.

The determinant of a skew-symmetric matrix $M$ of even dimension is the square of a polynomial, called the Pfaffian, in the entries of $M$.

As a corollary, this determinant is thus non-negative.

For example: $M=\begin{pmatrix} 0 & k \\ -k & 0 \end{pmatrix}$ be a skew-symmetric matrix.

Now $\det(M)=\begin{vmatrix} 0 & k \\ -k & 0 \end{vmatrix}=0-k(-k)=k^2$

so the Pfaffian is $k$.

In the $2×2$ case, then, any skew-symmetric matrix is nonsingular with a positive determinant, except the zero matrix.

Ref:

$1.$ https://www.quora.com/What-is-the-determinant-of-every-skew-symmetric-matrix

$2.$ http://mathworld.wolfram.com/Pfaffian.html